= Solution
Since $P_0=1$, part (d) gives $Q_0'=A_0/(1-x^2)$. A <partial fraction decomposition> and integration give, for $-1<x<1$,
$$
Q_0(x)=\frac{A_0}{2}\log\frac{1+x}{1-x}+B.
$$
The condition at zero fixes $B=0$. Thus the full conclusion from the stated condition is
$$
\boxed{Q_0(x)=A_0\operatorname{arctanh}x=\frac{A_0}{2}\log\frac{1+x}{1-x},\qquad A_0\ne0.}
$$
That condition alone does not fix the multiplicative constant. With the usual <Legendre function of the second kind> normalization $Q_0'(0)=1$, one has $A_0=1$. On real exterior intervals replace the logarithm by $\log|(1+x)/(1-x)|$; complex continuation requires a choice of logarithm branch.
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